Question

$\cos( \frac{7\pi}{12} ) =$
Va rog mult ​

Answer 1

$\bf cos\bigg(\dfrac{7\pi }{12} \bigg)= cos \bigg(\dfrac{4\pi +3\pi }{12} \bigg) =\\\bf cos\bigg(\dfrac{3\pi }{12}^{(3} + \dfrac{4\pi }{12}^{(4} \bigg) = cos \bigg( \dfrac{\pi }{4} + \dfrac{\pi }{3} \bigg) =\\\bf cos \bigg(\dfrac{\pi }{4} \bigg) \cdot cos \bigg(\dfrac{\pi }{3} \bigg) - sin \bigg( \dfrac{\pi }{4} \bigg) \cdot sin\bigg(\dfrac{\pi }{3} \bigg) =\\\bf =\dfrac{\sqrt{2} }{2} \cdot \dfrac{1}{2} -\dfrac{\sqrt{2} }{2} \cdot \dfrac{\sqrt{3} }{2} =\dfrac{\sqrt{2} }{4} - \dfrac{\sqrt{6} }{4} \bf \iff$

$\bf \iff \boxed{\bf \dfrac{\sqrt{2}-\sqrt{6} }{4} }$

$\bf \;$

$\;$$\bf sin(\, 30^{\circ} ) = \dfrac{1}{2}$

$\bf sin(\, 45^{\circ})=\dfrac{\sqrt{2} }{2}$

$\bf sin (\, 60^{\circ}) = \dfrac{\sqrt{3} }{2}$

$\bf \;$

$\bf cos(\, 30^{\circ})=\dfrac{\sqrt{3} }{2}$

$\bf cos(\, 45^{\circ})=\dfrac{\sqrt{2} }{2}$

$\bf cos(\, 60^{\circ})=\dfrac{1}{2}$

Answer 2

cos ( 7π / 12 ) = ?

= cos ( π / 4 + π / 3 )

= cos ( π / 4) cos ( π / 3 ) - sin ( π / 4 ) sin ( π / 3 )

= √2/2  * 1/2 - √2/2 - √3/2

= √2/2 * 1/2 - √2/2 * √3/2

= ( √2 - √6 ) / 4